What is the solution for y^{''}+2y^'+5y=0,y(0)=4,y^'(0)=-3 ?

The solution for y^{''}+2y^'+5y=0,y(0)=4,y^'(0)=-3 is y=e^{-t}(4cos(2t)+1/2 sin(2t))

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y^{''}+2y^'+5y=0,y(0)=4,y^'(0)=-3

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 4, y^{^{'}} (0) = - 3

y = e^{- t} (4 cos (2 t) + \frac{1}{2} sin (2 t))

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{- t} (4 cos (2 t) + \frac{1}{2} sin (2 t))

y = e^{- t} (4 cos (2 t) + \frac{1}{2} sin (2 t))

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What is the solution for y^{''}+2y^'+5y=0,y(0)=4,y^'(0)=-3 ?
The solution for y^{''}+2y^'+5y=0,y(0)=4,y^'(0)=-3 is y=e^{-t}(4cos(2t)+1/2 sin(2t))